Lecture Notes on Soundness and Correspondence

15-816: Modal Andre´ Platzer

Lecture 7 February 2, 2010

1 Introduction to This Lecture

In this lecture, we will cover the question how axiomatics and semantics of modal logic fit together. For correspondence theory results we refer to Schmitt [Sch03] and the book by Hughes and Cresswell [HC96].

2 Soundness

In the lecture 5 we have followed an axiomatic and a semantic approach to modal logic. But do these approaches fit together? We should not be using proof rules that make no sense semantically. Recall the axioms and proof rules we had so far Figure1. The proof rules are sound iff they can only derive semantical conse- quences. Note that, unlike in the first lectures, this is an external soundness notion. We do not justify one proof rule by checking compatibility with another proof rule (internal soundness). Instead, we check compatibility of all proof rules with respect to the external mathematical objects of the semantics.

Definition 1 (Soundness) A system S of proof rules and axioms of modal logic is (externally) sound iff, for all formulas ψ and all sets of formulas Φ:

Φ `S ψ implies Φ g ψ

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(P) all propositional tautologies

(K) (φ → ψ) → (φ → ψ)

(T) φ → φ

(4) φ → φ φ φ → ψ (MP) ψ φ (G) φ

Figure 1: Modal logic axioms

In particular, in order to check if a proof rule

φ1 . . . φn ψ is sound, we need to check for all instances of the proof rule if

φ1, . . . , φn g ψ In order to check if an axiom φ is sound, we need to check for all instances of the axiom if they are valid:

 φ

Lemma 2 AxiomK is sound.

Proof: Let K, s be a Kripke structure and a world. We have to prove that

K, s |= (φ → ψ) → (φ → ψ) Suppose the assumptions are true at s as there is nothing to show other- wise. Thus

K, s |= (φ → ψ) (1) K, s |= φ (2)

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We thus know for all s with sρt that K, t |= φ → ψ by (1) and also that K, t |= φ by (2). We have to show that K, s |= ψ. For that, we have to show for all t with sρt that K, t |= ψ. Consider any t with sρt. Since sρt, we know by (2) that K, t |= φ. Furthermore, since sρt, we know by (1) that K, t |= φ → ψ. Clearly, this implies K, t |= ψ. As t was arbitrary with sρt, this shows K, s |= ψ. 

Lemma 3 RuleG is sound.

Proof: We have to prove that φ g φ. Let K be a Kripke structure with K, t |= φ for all worlds t ∈ W . Now let s ∈ W be any world. We have to show that K, s |= φ. For that, let t ∈ W be any world with sρt. For t we have to show K, t |= φ. But this is simple, because, by assumption, K, t |= φ for all worlds t ∈ W . 

Note that the proof ofG crucially depends on knowing the premisses hold globally in all worlds. Thus, the notion of soundness that we are using here is also called global soundness, because it for the global consequence relation. Now when we try to prove soundness ofT, we run into problems. The rule just is not sound, although it looks so innocently helpful. What is wrong?

3 Correspondence

We have not been able to prove axiomT to be sound with respect to the semantics. This is a more systematic phenomenon that affects other rea- sonable modal axioms also, giving rise to the question how the axiomatic and semantic approach to modal logic correspond to one another. Correspondence theory tries to find connections between properties of Kripke frames and the formulas in modal logic that are true in all Kripke frames with this property.

Lemma 4 A Kripke frame (W, ρ) is reflexive K, s |= q → q for all Kripke structures K = (W, ρ, v). Here q is a propositional letter.

Proof: Assume that (W, ρ) is reflexive, i.e., (W, ρ)  ∀x ρ(x, x). Now let K = (W, ρ, v) be any Kripke structure, i.e., let v be any -value assign- ment for the Kripke frame (W, ρ). For very world s ∈ W , we have to show

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K, s |= q → q. Thus assume K, s |= q as there is nothing to show oth- erwise. Hence for all t with sρt we know K, t |= q. By reflexivity of ρ we know sρs, which implies K, s |= q. Conversely, let (W, ρ) be any Kripke frame that is not reflexive. Let r ∈ W be a world with (r, r) 6∈ ρ. We can choose any truth-assignment v for the Kripke frame (W, ρ). We choose to let ( true if rρs v(s)(q) := false otherwise

Consider the Kripke structure K = (W, ρ, v). Then K, r |= q but K, r 6|= q. Thus K, r 6|= q → q. 

This lemma says that the class of all reflexive frames is characterized by the formula q → q: Definition 5 (Characterization) Let C be a class of Kripke frames and φ a for- mula in modal logic. Formula φ characterizes C, if, for every Kripke frame (W, ρ):

(W, ρ) ∈ C iff for each v: K, s |= φ holds for K = (W, ρ, v)

Lemma 6 The formula q → q characterizes the class of all transitive frames.

Proof: Assume that K = (W, ρ, v) is any Kripke structure with a transitive Kripke frame (W, ρ). Consider any s ∈ W with K, s |= q as there is noth- ing to show otherwise. We have to show K, s |= q. Let t, r be any worlds with sρt and tρr. We have to show K, r |= q. By transitivity of ρ we know that sρt and tρr imply sρr. Because of K, s |= q, this implies K, r |= q. Conversely, let (W, ρ) be any Kripke frame that is not transitive. Let r0, r1, r2 ∈ W be worlds with r0ρr1 and r1ρr2 but (r0, r2) 6∈ ρ. We can choose any truth-assignment v for the Kripke frame (W, ρ). We choose ( true if r ρs v(s)(q) := 0 false otherwise

Consider the Kripke structure K = (W, ρ, v). Then K, r0 |= q but also K, r0 6|= q, because K, r2 6|= q. Thus K, r0 6|= q → q. 

Combining the above results obtained so far we have:

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Theorem 7 (Soundness of S4) The Kripke proof rules for S4 in Figure1 are sound for the class of reflexive and transitive frames.

Theorem 8 The conjunction of the following two multimodal formulas

ap → (p ∧ abp) a(p → bp) → (p → ap) characterizes the set of all multimodal Kripke frames (W, ρa, ρb) such that ρa is the reflexive, transitive closure of ρb.

Proof: If K = (W, ρa, ρb, v) is a Kripke structure where ρa is the reflexive, transitive closure of ρb, then it is easy to check that the two formulas are valid in K. Conversely, consider a Kripke frame (W, ρa, ρb) in which both formulas are valid, that is, for all Kripke structures (W, ρa, ρb, v). We have to show ∗ that ρa = ρb .

“⊆” Fix any s, t ∈ W with sρat. We show that (s, t) is in the reflexive, ∗ transitive closure ρb of ρb. We choose the valuation map ( 1 if (s, w) ∈ ρ ∗ v(w)(p) = b 0 otherwise

First let us show for K = (W, ρa, ρb, v) that the assumption holds:

K, s |= a(p → bp)

For that, let w ∈ W be any world with sρaw and K, w |= p. According ∗ to our choice of v, this implies that (s, w) ∈ ρb . Now for any world 0 0 0 ∗ w ∈ W with wρbw , we note that (s, w ) ∈ ρb by the definition of a transitive closure. Thus, our choice of v ensures K, w0 |= p. Conse- quently, K, w |= bp. Now we have shown K, s |= a(p → bp). We assumed that a(p → bp) → (p → ap)

is valid in K, thus K, s |= p → ap. Since the transitive closure is reflexive, our choice of v implies that K, s |= p, hence K, s |= ap. But the world t we fixed in the beginning was one of the worlds with sρat, ∗ hence K, t |= p. Now our choice of v implies that (s, t) ∈ ρb is in the ∗ reflexive, transitive closure. In summary, ρa ⊆ ρb .

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∗ “⊇” Fix any s, t ∈ W with (s, t) ∈ ρb , i.e., that is in the reflexive, transitive closure. We want to show that sρat. We choose the valuation map ( 1 if sρ w v(w)(p) = a 0 otherwise

By premiss K, s |= ap → (p ∧ abp) for K = (W, ρa, ρb, v). Accord- ing to our choice of v, we have the assumption K, s |= ap. Thus the premiss implies K, s |= p ∧ abp (3) ∗ We have assumed (s, t) ∈ ρb . So there are worlds w0, w1, . . . , wn such that wiρbwi+1 for all i < n and w0 = s and wn = t. Let us show by induction on i that K, wi |= p.

0. For i = 0 this is implied by (3).

i. Assume K, wi |= p. According to our choice of v this implies that sρawi, hence (3) implies K, wi |= bp and K, wi+1 |= p.

Finally for wn = t this implies K, t |= p, which, by our choice of v ∗ implies sρat. In summary, ρa ⊇ ρb .



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Exercises

Exercise 1 Give correspondence results for the following cases. That is, for each modal formula identify the class of frames that it characterizes. Explain why and prove correspondence, or explain why there is no correspondence at all.

1. q → ♦q

2. ♦q → ♦q

3. ♦q ↔ q

4. (a → b) → (b → a)

Exercise 2 Give correspondence results for the following cases. That is, for each class of frames find a modal formula that characterizes it. Explain why and prove correspondence, or explain why there is no correspondence at all.

1. The class of all symmetric frames.

2. The class of all dense frames, i.e., if sρt then sρz, zρt for some z.

3. The class of all frames where sρt, sρu imply tρz, uρz for some z.

Exercise 3 Show correspondence of (a ∧ a → b) ∨ (b ∧ b → a) (weakly connected = sRt and sRu imply tRu or uRt or t=u) with the class of all frames where sρt, sρu imply tρu or uρt or t = u.

Exercise 4 Describe a general correspondence result between modal logic formu- las and classes of Kripke frames. That is, to each class of Kripke frames, give one formula in modal logic that characterizes it. Explain!

Exercise 5 Prove or disprove: If A and B be two formulas in propositional modal logic that characterize the same class of Kripke structures, then A ↔ B is valid.

Exercise 6 After Definition1, variations of soundness have been given for proof rules and for axioms. Do these notions coincide with the soundness actually de- fined in Definition1? Prove or disprove.

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References

[HC96] G.E. Hughes and M.J. Cresswell. A New Introduction to Modal Logic. Routledge, 1996.

[Sch03] Peter H. Schmitt. Nichtklassische Logiken. Vorlesungsskriptum Fakultat¨ fur¨ Informatik , Universitat¨ Karlsruhe, 2003.

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